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Evaluate each of the following:

\(sin^{-1}(sin\frac{\pi}6)\)

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The value of sin\(\frac{\pi}6\) is \(\frac 12\)

∴ The question becomes sin–1\((\frac 12)\)

The range of principal value of sin–1 is \([\frac{-\pi}2,\frac{\pi}2] \) and sin\((\frac{π}6)\) = \(\frac 12\)

Therefore, the value of sin–1(sin\(\frac{\pi}6\)) is \(\frac{\pi}6.\)

sin–1(sin x) = x

Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2] \)

∴ we can write sin–1(sin\(\frac{\pi}6\)) = \(\frac{\pi}6\)

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